Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #61

Solve by the matrix method: (1)x + (-1)y + (2)z = 7; (2)x + (-1)y + (3)z = 12; (3)x + (2)y + (-1)z = 5.

✅ Step-by-Step Solution

Final Answer: x=2, y=1, z=3

Write MX=B, where M=\(\begin{bmatrix}1 & -1 & 2\\2 & -1 & 3\\3 & 2 & -1\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}7\\12\\5\end{bmatrix}\).
The determinant is -2≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-5 & 11 & 7\\3 & -7 & -5\\-1 & 1 & 1\end{bmatrix}\).
Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-5 & 3 & -1\\11 & -7 & 1\\7 & -5 & 1\end{bmatrix}\), hence M⁻¹=adj(M)/(-2)=\(\begin{bmatrix}5/2 & -3/2 & 1/2\\-11/2 & 7/2 & -1/2\\-7/2 & 5/2 & -1/2\end{bmatrix}\).
Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}2\\1\\3\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
Therefore x=2, y=1, z=3.

[1 mark] Write MX=B, where M=\(\begin{bmatrix}1 & -1 & 2\\2 & -1 & 3\\3 & 2 & -1\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}7\\12\\5\end{bmatrix}\).
[1 mark] The determinant is -2≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}-5 & 11 & 7\\3 & -7 & -5\\-1 & 1 & 1\end{bmatrix}\).
[1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}-5 & 3 & -1\\11 & -7 & 1\\7 & -5 & 1\end{bmatrix}\), hence M⁻¹=adj(M)/(-2)=\(\begin{bmatrix}5/2 & -3/2 & 1/2\\-11/2 & 7/2 & -1/2\\-7/2 & 5/2 & -1/2\end{bmatrix}\).
[1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}2\\1\\3\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
[1 mark] Therefore x=2, y=1, z=3.

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